Question: Let $y=\dfrac{x^2}{\cos(x)}$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{2x\cos(x)+x^2\sin(x)}{\cos^2(x)}$ (Choice B) B $-\dfrac{2x}{\sin(x)}$ (Choice C) C $\dfrac{2x-\sin(x)}{\cos^2(x)}$ (Choice D) D $\dfrac{2x\cos(x)-x^2\sin(x)}{\cos(x)}$
$\dfrac{x^2}{\cos(x)}$ is the quotient of two, more basic, expressions: $x^2$ and $\cos(x)$. Therefore, $\dfrac{dy}{dx}$ can be found using the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\dfrac{x^2}{\cos(x)}\right) \\\\ &=\dfrac{\dfrac{d}{dx}(x^2)\cos(x)-x^2\dfrac{d}{dx}(\cos(x))}{(\cos(x))^2}&&\gray{\text{The quotient rule}} \\\\ &=\dfrac{2x\cdot \cos(x)-x^2\cdot (-\sin(x))}{(\cos(x))^2}&&\gray{\text{Differentiate }x^2\text{ and }\cos(x)} \\\\ &=\dfrac{2x\cos(x)+x^2\sin(x)}{\cos^2(x)}&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{2x\cos(x)+x^2\sin(x)}{\cos^2(x)}$